Question

$\int \frac{dx}{\sqrt{2ax-x^2}}=a^n{\sin }^{-1}[\frac{x}{a}-1]$
The value of n is : ?

• A. $0$
• B. $-1$
• C. $1$
• D. None of these

Answer 1

The correct option is A $0$

$\int \frac{dx}{\sqrt{2ax-x^2}}$

$=\int \frac{dx}{\sqrt{a^2-(a^2-2ax+x^2)}}$

$=\int \frac{dx}{\sqrt{a^2-{(x-a)}^2}}$

Let $t=x-a\Rightarrow dt=dx$

$=\int \frac{dt}{\sqrt{a^2-t^2}}$

Let $t=a\sin \theta \Rightarrow dt=a\cos \theta d\theta$

$=\int \frac{a\cos \theta d\theta }{\sqrt{a^2-a^2{\sin }^2\theta }}$

$=\int \frac{a\cos \theta d\theta }{a\cos \theta }$

$=\int d\theta$

$=\theta +c$

$={\sin }^{-1}\frac{t}{a}+c$ using $t=a\sin \theta$

$={\sin }^{-1}\frac{x-a}{a}+c$ using $t=x-a$

$=1\times {\sin }^{-1}(\frac{x}{a}-1)+c$

Comparing with $a^n{\sin }^{-1}(\frac{x}{a}-1)$ we have

$a^n=1=a^0$

$\Rightarrow n=0$

$\therefore$ the value of $n=0$